Parabola
Parabola
nta_pyq_2025_apr
Grade 11
Question:
Let $P(4, 4\sqrt{3})$ be a point on the parabola $y^2 = 4ax$ and $PQ$ be a focal chord of the parabola. If $M$ and $N$ are the foot of perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola, then the area of the quadrilateral $PQNM$ is equal to
$17\sqrt{3}$
$\dfrac{263\sqrt{3}}{8}$
$\dfrac{34\sqrt{3}}{3}$
$\dfrac{343\sqrt{3}}{8}$
Step-by-Step Solution
Key Concept: $PQNM$ is a trapezium with $PM\parallel QN$ (both perpendicular to the directrix); $|MN|=|y_P-y_Q|$ and $|PQ|=SP+SQ$ (focal radii); area $=\tfrac{1}{2}|MN|\cdot|PQ|$.
$P=(4,4\sqrt{3})$ on $y^2=4ax$: $48=16a \Rightarrow a=3$, parabola $y^2=12x$. Parameter of $P$: $2at_1=4\sqrt{3} \Rightarrow t_1=\tfrac{2}{\sqrt{3}}$. Focal chord: $t_1t_2=-1 \Rightarrow t_2=-\tfrac{\sqrt{3}}{2}$, so $Q=\left(\tfrac{9}{4},-3\sqrt{3}\right)$. $SP=a(1+t_1^2)=3(1+\tfrac{4}{3})=7$; $SQ=a(1+t_2^2)=3(1+\tfrac{3}{4})=\tfrac{21}{4}$; $|PQ|=SP+SQ=\tfrac{49}{4}$. $|MN|=|y_P-y_Q|=|4\sqrt{3}+3\sqrt{3}|=7\sqrt{3}$. Area $=\tfrac{1}{2}\cdot7\sqrt{3}\cdot\tfrac{49}{4}=\dfrac{343\sqrt{3}}{8}$.
Correct Answer: 4